Respuesta :
Answer:
x=-6 and x=1
Step-by-step explanation:
Given that:
[tex]3x^2+15x-18[/tex]
Now,
[tex]3x^2+15x-18=0\\[/tex]
Using the quadratic formula where a=3, b=15, and C=-18
x=-b±[tex]\sqrt{b^2-4ac} /2a[/tex]
x=-15±[tex]{\sqrt{15^2-4(3)(-18} } /2(3)[/tex]
x=-15±[tex]\sqrt{441} /6[/tex]
the discriminant [tex]b^2-4ac>0[/tex]
So, there are two real roots
x=-15±21/6
[tex]x=\frac{6}{6} \\x=1\\x=\frac{-36}{6} \\x=-6[/tex]
Therefore, x=-6 and x=1
